🟩 The story of Wordle

The Wordle is a web-based word game created and developed by Welsh software engineer Josh Wardle.
It has become incredibly popular, thanks to its simplicity, playability and challenge… and also because players around the world can show off their prowess on by posting their daily results on social media as emoji squares.

The aim of the game is to guess a five-letter word using a maximum of six attempts.
There is a unique “wordle” every day and the rules are simple.
After every guess you receive feedback for each letter in your guess in the form of a coloured square:
⬛ means the that the letter is not in the word
🟨 means the letter is in the word but in a different position to the one you chose.
🟩 means the letter is in the word in the position you chose.

This Assessment asks you to investigate the results posted by a sample of Australian and international players.

What makes a word hard to guess?
How many guesses does it take to crack the puzzle?
And, critically, which Australian state or territory can boast the best wordlers?

…your data awaits…

Instructions

📢 About this Assessment item

This individual assignment is designed to evaluate, among other things, your abilities to
- apply your data manipulation knowledge and skills on real datasets
- read, interpret and follow instructions about working with data
- identify and explain critical ethical issues in data science scenarios
- look things up and figure things out.
  - Note that Blackboard/References provides advice and useful references for looking things up
Details about the weighting and due date can be found at Blackboard/Assessment02
We have attempted to make this Assessment perfect
- However, because we are running it for the first (and last!) time we may need to make corrections
- Any corrections or clarifications will be published in Blackboard/Assessment02/Errata

📩 How to submit this Assessment item

Question 6 provides instructions as to how to submit this Assessment item.
- Correctly following these instructions is worth 1 mark.

😎 How to succeed with this Assessment item

We expect you to answer all questions.
- Remember, the more you give us, the more opportunity we have to award marks
- If you are encountering difficult circumstances, here is information about applying for an extension
  - Note that it is QUT (not the DSB100 Teaching Team) who assesses extension requests
We will release your marks on Blackboard as soon as we have received, marked and moderated all submissions, including extensions

🛠️ Work in a new directory, using relative file names

Uncompress your Assessment files in a newly created directory
Refer to files using names relative to the location of this R markdown document, e.g., ../data,
- This will mean that the Teaching Team can run your submission in a new location

🏡 Work from home!

Don’t change your working directory
- This helps ensure that your code can be run without necessarily having to start from the top

🧹 Start with a clean environment

It is good practice to make sure that our code runs in a “clean” environment
- Here “clean” means free from R objects that are older than the code we are working on
- Working in a clean environment means that our code runs from a well-defined initial state
- If we forget to do this, it is easy to create code that relies on objects that we have left lying around
  - This code will fail when we give our code to someone else to run in their clean environment
One simple way to ensure we start in a clean environment is to remove everything other than the code files we are working on:

rm(list=ls()) # Remove all objects

Notice that we’ve put code and output that we provide for you in green-shaded text boxes

📚 Load only these libraries

This assessment can and should be completed using only the following four libraries.

Do not use any other libraries

library('ggplot2')
library('dplyr') 
library('lubridate')
library('plotly')

🟨 Your datasets

This assessment uses two datasets of tweet data from wordle results which we load here:

wordle.int <- readRDS('../data/wordle.int.RDS') # A sample of 20000 wordle results from international players

Warning in readRDS("../data/wordle.int.RDS"): strings not representable in native encoding will be translated to UTF-8

wordle.aus <- readRDS('../data/wordle.aus.RDS') # A sample of 11389 wordle results from Australian players

Both wordle.int and wordle.aus contain the variables

wordle_id: a unique consecutive identifier of the wordle.
solution: the solution or wordle of the day: e.g. POWER
date: date of the published result by the user on Twitter.
attempts: the number of attempts (out of 6) used to solve the wordle.
score: the score achieved, where each ⬛ = 2, 🟨 = 1 and 🟩 = 0. The smaller the better.
vowels: number of vowels in the solution. It affects the difficulty of the guess. The Y is considered a vowel here.
repeated: consecutive / double letters in the solution e.g. SWEET. It affects the difficulty of the guess.
freq: the log of the n-gram mean frequencies of the word (during the last 5 years).
- This was obtained from Google Ngram Viewer website.
- We expect that more frequently used words should be easier to guess.
wday: day of the week (Mon, Tue, …, Sun).
tweet: a tweet spoiler-free emoji grid published by the users.

The Australian dataset, wordle.aus also contain the variable

city: the city of the Twitter user.

Click to show a glimpse of the international wordle data:

Rows: 20,000
Columns: 11
Rowwise: 
$ wordle_id  <dbl> 386, 364, 234, 249, 231, 357, 327, 268, 279, 269, 389, 381, 412, 221, 372, 367, 380, 341, 230, 282,~
$ solution   <chr> "BERTH", "CACAO", "FRAME", "TROVE", "ALOFT", "GOOSE", "SLUNG", "SMELT", "DEPOT", "TEASE", "BLAND", ~
$ date       <date> 2022-07-10, 2022-06-18, 2022-02-08, 2022-02-23, 2022-02-05, 2022-06-11, 2022-05-12, 2022-03-14, 20~
$ attempts   <dbl> 5, 6, 5, 5, 4, 5, 6, 3, 4, 2, 3, 5, 6, 3, 4, 3, 3, 3, 4, 2, 4, 4, 3, 4, 2, 5, 5, 3, 4, 6, 4, 6, 2, ~
$ score      <dbl> 26, 34, 18, 16, 11, 33, 26, 15, 21, 7, 8, 28, 33, 6, 22, 12, 11, 12, 17, 7, 15, 18, 16, 14, 7, 24, ~
$ vowels     <int> 1, 3, 2, 2, 2, 3, 1, 1, 2, 3, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 3, 2, 2, 1, 2, 2, ~
$ repeated   <dbl> 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ~
$ freq       <dbl> -19.21077, -17.90610, -15.74757, -19.01154, -19.23042, -16.46714, -19.97755, -19.25412, -17.31333, ~
$ wday       <ord> Sun, Sat, Tue, Wed, Sat, Sat, Thu, Mon, Fri, Tue, Wed, Tue, Fri, Wed, Sun, Tue, Mon, Thu, Fri, Mon,~
$ tweet      <chr> "\n\n<U+2B1B><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E8>\n<U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E8><U+2B1B>\n<U+2B1B><U+0001F7E9><U+0001F7E8><U+2B1B><U+0001F7E8>\n<U+0001F7E8><U+0001F7E9><U+0001F7E8><U+2B1B><U+2B1B>\n<U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9>", "\n\n<U+2B1B><U+0001F7E8><U+2B1B><U+2B1B><U+2B1B>\n<U+0001F7E9><U+2B1B><U+0001F7E8><U+2B1B><U+2B1B>\n<U+0001F7E9><U+0001F7E9>~
$ tweet_text <chr> "Wordle 386 5/6\n\n<U+2B1B><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E8>\n<U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E8><U+2B1B>\n<U+2B1B><U+0001F7E9><U+0001F7E8><U+2B1B><U+0001F7E8>\n<U+0001F7E8><U+0001F7E9><U+0001F7E8><U+2B1B><U+2B1B>\n<U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9>", "Wordle 364 6/6\n\n~

<U+0001F7E8><U+0001F7E9><U+0001F7E9><U+0001F7E8><U+0001F7E8><U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E8><U+0001F7E8><U+2B1B><U+0001F7E8><U+2B1B><U+0001F7E8><U+0001F7E9><U+0001F7E8><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E8><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E8><U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E8><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E9><U+2B1B><U+2B1B><U+0001F7E9><U+0001F7E9><U+2B1B><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E8><U+0001F7E9><U+0001F7E9><U+0001F7E9>

❔ Question 1: Summarizing wordle data (8 marks total)

This question asks you to summarise and explore the wordle.int dataset, a sample of 20,000 tweets about wordle from around the world. This kind of summarisation is a critical step in understanding any dataset.

Q1.1

Demonstrate R commands to generate summary statistics for each wordle in the wordle.int dataset:

group the data by wordle, solution, vowels, repeated, freq, and wday
count the numbers of non-missing values for the score and for the attempts variable.
compute the mean of score and attempts
compute the standard deviation of score and attempts
save your results to wordle.int.summary and print the first 10 rows of that object
your results should look like this:

(2 marks)

# A tibble: 216 x 12
# Groups:   date, solution, vowels, repeated, freq [216]
   date       solution vowels repeated  freq wday  score_n score_mean score_sd attempts_n attempts_mean attempts_sd
   <date>     <chr>     <int>    <dbl> <dbl> <ord>   <int>      <dbl>    <dbl>      <int>         <dbl>       <dbl>
 1 2022-01-15 PANIC         2        0 -16.5 Sat        67       18.5     7.39         67          3.78        7.39
 2 2022-01-16 SOLAR         2        0 -15.7 Sun        60       17.4     6.83         60          4.03        6.83
 3 2022-01-17 SHIRE         2        0 -18.7 Mon        45       13.3     7.29         45          3.44        7.29
 4 2022-01-18 PROXY         2        0 -17.6 Tue        50       25.1     8.31         50          4.72        8.31
 5 2022-01-19 POINT         2        0 -14.0 Wed        55       17.2     6.01         55          3.55        6.01
 6 2022-01-20 ROBOT         2        0 -16.6 Thu        57       20.1     8.82         57          3.91        8.82
 7 2022-01-21 PRICK         1        0 -18.6 Fri       145       18.3     7.29        145          3.85        7.29
 8 2022-01-22 WINCE         2        0 -20.8 Sat        59       21.8     7.99         59          4.39        7.99
 9 2022-01-23 CRIMP         1        0 -19.3 Sun       102       18.7     6.97        102          3.95        6.97
10 2022-01-24 KNOLL         1        1 -19.1 Mon        72       26.3     6.68         72          4.54        6.68
# … with 206 more rows

wordle.int %>%
  group_by(date, wordle_id, solution, vowels, repeated, freq, wday) %>%
  summarise(score_n = sum(!is.na(score)), score_mean = mean(score), score_sd = sd(score), attempts_n = sum(!is.na(attempts)), attempts_mean = mean(attempts), attempts_sd = sd(attempts)) -> wordle.int.summary

head(wordle.int.summary, 10)

# A tibble: 10 x 13
# Groups:   date, wordle_id, solution, vowels, repeated, freq [10]
   date       wordle_id solution vowels repeated  freq wday  score_n score_mean score_sd attempts_n attempts_mean
   <date>         <dbl> <chr>     <int>    <dbl> <dbl> <ord>   <int>      <dbl>    <dbl>      <int>         <dbl>
 1 2022-01-15       210 PANIC         2        0 -16.5 Sat        67       18.5     7.39         67          3.78
 2 2022-01-16       211 SOLAR         2        0 -15.7 Sun        60       17.4     6.83         60          4.03
 3 2022-01-17       212 SHIRE         2        0 -18.7 Mon        45       13.3     7.29         45          3.44
 4 2022-01-18       213 PROXY         2        0 -17.6 Tue        50       25.1     8.31         50          4.72
 5 2022-01-19       214 POINT         2        0 -14.0 Wed        55       17.2     6.01         55          3.55
 6 2022-01-20       215 ROBOT         2        0 -16.6 Thu        57       20.1     8.82         57          3.91
 7 2022-01-21       216 PRICK         1        0 -18.6 Fri       145       18.3     7.29        145          3.85
 8 2022-01-22       217 WINCE         2        0 -20.8 Sat        59       21.8     7.99         59          4.39
 9 2022-01-23       218 CRIMP         1        0 -19.3 Sun       102       18.7     6.97        102          3.95
10 2022-01-24       219 KNOLL         1        1 -19.1 Mon        72       26.3     6.68         72          4.54
# ... with 1 more variable: attempts_sd <dbl>

Note In case you cannot complete Q1.1, we now provide wordle.int.summary to be used from here on.

wordle.int.summary <- readRDS("../data/wordle.int.summary.RDS")

Q1.2

Demonstrate R commands to show the proportion of wordles in wordle.int.summary containing different numbers of vowels , rounded to 3 decimal places, ensuring that any missing values (NAs) are included.

(1 mark)

# Write your answer here

table(wordle.int.summary$vowels, useNA = "ifany")


  1   2   3   4 
 50 136  29   1

# We can use this table to obtain proportions by dividing each value by the total number of wordles in the data set. E.g. 
totalVowels = 50 + 136 + 29 + 1 # Values taken from above table function.
oneVowel = round((50/totalVowels), 3)
twoVowels = round((136/totalVowels), 3)
threeVowels = round((29/totalVowels), 3)
fourVowels = round((1/totalVowels), 3)

cat("Proportion of wordles containing one vowel: ", oneVowel, "\nContaining two vowels: ", twoVowels,
    "\nContaining three vowels: ", threeVowels, "\nContaining four vowels", fourVowels)

Proportion of wordles containing one vowel:  0.231 
Containing two vowels:  0.63 
Containing three vowels:  0.134 
Containing four vowels 0.005

Q1.3

Demonstrate R commands to show that there are 216 different days of results in wordle.int.summary

(1 mark)

# Write your answer here
n_distinct(wordle.int.summary$date)

[1] 216

Q1.4

Demonstrate R commands that add a month variable to wordle.int.summary so you can answer the following question.

In July 2022, what was the hardest word, based on the mean number of attempts in wordle.int.summary
(i.e., the more attempts the harder)? * Hints: * Use lubridate’s month() function. * wordle.int.summary is a grouped data frame * Depending on your approach, you may have to remove the grouping to answer these questions

(1 mark)

# Write your answer here
wordle.int.summary <- wordle.int.summary %>%
  mutate(month = month.abb[lubridate::month(date)])

View(wordle.int.summary) # We manually view the data set to see which word in July 2022 has the highest mean number of attempts

print("Based on the above commands, the hardest word in July 2022 is 'Cinch' on the 26th, with a mean attempt of 4.55...")

[1] "Based on the above commands, the hardest word in July 2022 is 'Cinch' on the 26th, with a mean attempt of 4.55..."

Q1.5

Using wordle.int.summary, demonstrate R commands that compute and display the following summary statistics, by month, for the attempts_mean variable:
- mean and the median.
- standard deviation.
- quartiles Q1 and Q3.
  (1 mark)

Note We have asked you to summarise the mean number of attempts of each wordle: in effect, you are providing summaries of a summary statistic. The results that you will get will be slightly different than if you were to work with the individual observations (i.e., the integer number of attempts in wordle.int). We have done this to avoid overplotting in Q1.6.

# Write your answer here
wordle.int.summary %>%
  group_by(month) %>%
  summarize(mean = mean(attempts_mean), median = median(attempts_mean), standard_dev = sd(attempts_mean), Q1 = quantile(attempts_mean, 0.25), 
            Q3 = quantile(attempts_mean, 0.75))

# A tibble: 8 x 6
  month  mean median standard_dev    Q1    Q3
  <chr> <dbl>  <dbl>        <dbl> <dbl> <dbl>
1 Apr    4.16   4.14        0.356  3.99  4.46
2 Aug    4.28   4.28        0.381  4.08  4.49
3 Feb    4.14   4.16        0.346  3.93  4.33
4 Jan    4.05   4.02        0.331  3.89  4.25
5 Jul    4.19   4.20        0.234  3.99  4.37
6 Jun    4.15   4.09        0.334  3.94  4.28
7 Mar    4.15   4.19        0.311  3.94  4.32
8 May    4.14   4.12        0.314  3.96  4.29

Q1.6

Demonstrate R commands create violin plots of attempts_mean for each week day of the wordle.int.summary data

draw the 0.25, 0.5 and 0.75 quantiles on the violins
overlay these with blue, semi-transparent, horizontally jittered points (width=0.1)
give the plot
- the title “International Wordle data”
- the subtitle “2022-01-15 to 2022-08-21”
- no x-axis label
- the y-axis label “daily mean number of attempts”
ensure the y axis ticks are the counting numbers from 1 to 6
(1 mark)

# Write your answer here
ggplot(data=wordle.int.summary, aes(x = wday, y = attempts_mean)) +
  geom_violin() +
  geom_boxplot() +
  geom_jitter(height = 0.05, width = 0.1, colour = "blue", alpha = 0.15) +
  labs(title = "International Wordle data",
       subtitle = "2022-01-15 to 2022-08-21",) +
  xlab("") +
  scale_y_continuous(name = "daily mean number of attempts",
                     limits = c(1,6),
                     breaks = 1:7,
                     expand = c(0,0))

Q1.7

Consider the summaries that you have been calculating. Ask yourself: how confident are you that they are based on correct data?

In practice (in contrast to this assessment task) our first step in working with data should be to explore its values to make sure they are reasonable.
We have left that to the last part of Question 1 so you can see what can go wrong if we don’t do that first.

Consider the following plot and the code that generated it.

ggplot(wordle.int, aes(x=factor(attempts))) + 
  geom_bar() + 
  facet_grid(month(date, label = TRUE)~.) + 
  labs(
    title="Monthly summaries of wordle attempts",
    subtitle="International wordle data",
    x="number of attempts")

Comment on what this graph shows, what it implies for your summaries of the number of attempts, and what you would do in light of this.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

The number of attempts monthly roughly follow a normal distribution. It confirms previous summaries of number of attempts that have the mean revolve around 4. Though it may also confirm outliers in the data. We could fix this by potentially removing the data point, assign the next value nearer to the median in place of the outlier value, etc.

<U+0001F7E8><U+0001F7E9><U+2B1B><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+0001F7E9><U+2B1B><U+0001F7E9><U+0001F7E9><U+0001F7E8><U+0001F7E9><U+0001F7E9><U+0001F7E8><U+2B1B><U+2B1B><U+0001F7E9><U+0001F7E9><U+2B1B><U+0001F7E8><U+0001F7E8><U+0001F7E8><U+2B1B><U+0001F7E8><U+2B1B><U+2B1B><U+0001F7E9><U+2B1B><U+2B1B><U+0001F7E8><U+0001F7E8><U+2B1B><U+0001F7E9><U+2B1B><U+2B1B><U+0001F7E9><U+0001F7E8><U+0001F7E9><U+0001F7E8><U+0001F7E8><U+2B1B><U+0001F7E9><U+2B1B><U+2B1B><U+0001F7E8>

❔ Question 2: Exploring Australian wordles (6 marks total)

The sample of Australian wordles in wordle.aus was created by selecting tweets whose geolocation showed they originated within a 40km radius of state and territory capitals.

Q2.1

Demonstrate R commands that show the mean and standard deviation of score and attempts by city in the Australian wordle dataset (wordle.aus)

(1 mark)

# Write your answer here
wordle.aus %>%
  group_by(city) %>%
  summarize(mean_score = mean(score), sd_score = sd(score), mean_attempts = mean(attempts), sd_attempts = sd(attempts))

# A tibble: 8 x 5
  city      mean_score sd_score mean_attempts sd_attempts
  <chr>          <dbl>    <dbl>         <dbl>       <dbl>
1 adelaide        19.1     7.51          4.09       1.06 
2 brisbane        19.3     7.42          4.11       1.09 
3 canberra        19.2     7.39          4.10       1.08 
4 darwin          19.0     6.29          3.96       0.903
5 hobart          19.5     7.42          4.04       1.02 
6 melbourne       19.5     7.28          4.14       1.07 
7 perth           19.4     7.44          4.11       1.10 
8 sydney          19.7     7.16          4.15       1.05

Here we provide a more detailed data frame with attempts and score summarised by city per date.

# A tibble: 240 x 10
# Groups:   city, date, freq [240]
   city     date    freq solution score_n score_mean score_sd attampts_n attempts_mean attempts_sd
   <chr>    <date> <dbl> <chr>      <int>      <dbl>    <dbl>      <int>         <dbl>       <dbl>
 1 adelaide NA     -22.7 COYLY         25       26.6     5.53         25          5.24       0.831
 2 adelaide NA     -20.3 HUNKY         29       23.3     6.17         29          4.31       1.07 
 3 adelaide NA     -20.1 ELOPE         31       20       7.14         31          4.16       0.969
 4 adelaide NA     -20.0 GRUEL         27       23       6.10         27          4.52       0.893
 5 adelaide NA     -19.9 CINCH         27       22.4     5.78         27          4.48       0.975
 6 adelaide NA     -19.9 GLEAN         28       17.5     7.32         28          4          1.22 
 7 adelaide NA     -19.8 SHRUG         30       18.2     6.43         30          4          1.05 
 8 adelaide NA     -19.3 TWANG         30       17.3     6.74         30          3.73       0.785
 9 adelaide NA     -19.3 UNFIT         34       20.4     6.04         34          4.12       1.04 
10 adelaide NA     -19.3 KHAKI         29       23.7     6.63         29          4.31       0.850
# ... with 230 more rows

Q2.2

Using the data frame wordle.aus.city.date and ggplot(), create an interactive visualisation of the relationship between attempts_mean and score_mean with a scatter plot.

Show data from each city using points with a different color and area (not radius) proportional to size_n
- Use an alpha of 0.5 to deal with overplotting
Add a weighted linear fit to the graph
- Map score_n to the weight aesthetic
- Hint: look at the geom_smooth() function
Use the function ggplotly in the package plotly to produce an interactive graph.
(2 marks)

# Write your answer here
ggplot(data=wordle.aus.city.date, aes(x=attempts_mean, y=score_mean)) +
  geom_point(alpha=0.5, aes(colour=factor(city), size=score_n)) +
  geom_smooth(method="lm", aes(weight=score_n)) -> Q2.2

ggplotly(Q2.2)

Q2.3

Using wordle.aus.city.date, demonstrate code to:

Create a vector of the mean attempts for Sydney and a vector of the mean attempts for Canberra.
(1 mark)

# Write your answer here
syd_mean_attempts <- wordle.aus.city.date$attempts_mean[wordle.aus.city.date$city=="sydney"]
canb_mean_attempts <- wordle.aus.city.date$attempts_mean[wordle.aus.city.date$city=="canberra"]

Using the vectors you have just created, perform a statistical test to assess whether Canberra and Sydney have similar performance in terms of mean attempts.
(1 mark)

# Write your answer here
t.test(syd_mean_attempts, canb_mean_attempts)


    Welch Two Sample t-test

data:  syd_mean_attempts and canb_mean_attempts
t = 0.44163, df = 52.241, p-value = 0.6606
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.1640305  0.2566177
sample estimates:
mean of x mean of y 
 4.163191  4.116898

Interpret the outcome of your statistical test

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

The t-test tests two hypotheses, being the null hypothesis (no difference between the true averages of the two groups) versus the alternate hypothesis (a difference between the true averages of the two groups). The p-value of 0.6606 is the probability of getting a t-statistic whose magnitude is equal to or greater than 0.44163 (test statistic), with 52.2 degrees of freedom. This p-value means that we cannot reject the null hypothesis. Based on this, we can say that there is not enough evidence of a difference between the two averages of the two groups at the usual significance level of 0.05.

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❔ Question 3: Linear regression (5 marks total)

This question explores two different linear regression models of the wordle.int.summary data to predict attempts_mean.

Note that, since the response variable is based on counts, Poisson regression would be more appropriate.
However, since we have not covered the fitting and interpretation of Poisson regression, we will stick to linear regression in this instance

Use the following covariates or independent variables to predict the response (attempts_mean):

model 1: freq, vowels and repeated.
model 2: freq and vowels .

Q3.1

Demonstrate how to fit both linear regression models.

(1 mark)

# Write your answer here   
model_1.lm <- lm(attempts_mean ~ freq + vowels + repeated, data = wordle.int.summary)
model_2.lm <- lm(attempts_mean ~ freq + vowels, data = wordle.int.summary)

model_1.lm


Call:
lm(formula = attempts_mean ~ freq + vowels + repeated, data = wordle.int.summary)

Coefficients:
(Intercept)         freq       vowels     repeated  
     2.9893      -0.0569       0.0701       0.2543

model_2.lm


Call:
lm(formula = attempts_mean ~ freq + vowels, data = wordle.int.summary)

Coefficients:
(Intercept)         freq       vowels  
    3.04380     -0.05673      0.05793

Q3.2

Examine the coefficients of model 1 and explain in general terms what this suggests as the n-gram frequency of a wordle or the number of vowels in a wordle increase.

(1 mark)

# Write your code here   
model_1.lm$coefficients

(Intercept)        freq      vowels    repeated 
 2.98926510 -0.05690313  0.07010316  0.25432926

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

For every 1 unit of frequency increase, the expected mean attempts lowers by 0.057. For every 1 unit of number of vowels increase in a wordle, the expected mean attempts increases by 0.70.

Q3.3

In terms of Akiake’s Information Criterion, explain which model better fits the data.

(1 mark)

# Write your answer here   
AIC(model_1.lm)

[1] 90.86752

AIC(model_2.lm)

[1] 104.3646

Model 1 better fits the data. AIC penalizes models that use more parameters. Both models explain the same variation, and since model 1 has lower AIC, it has fewer parameters and thus will have a better fit model.

Q3.4

Interpret the apparent effect of repeated characters on attempts_mean in Model 1.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

From Q3.2, we can see that the repeated predictor in model 1 has a coefficient of 0.254. This means that for every 1 unit increase in repeated characters, the expected mean attempts increases by 0.254.

Q3.5

Demonstrate code to visually assess whether the residuals in Model 1 are normally distributed and briefly discuss what you conclude and why?

Hint: plot.lm plots a range of diagnostics for lm objects
(1 mark)

# Write your answer here   
plot(model_1.lm)

# See workshop 9.1

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

The Normal Q-Q plot seems to indicate a normal distribution. The circles within the plot seem to follow the normality line well.

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❔ Question 4: Logistic regression (6 marks total)

Here, we create an additional variable called difficulty in the dataframe wordle.int.summary

if attempts_mean $\leq 4.1$ then difficulty is low
if attempts_mean $> 4.1$ then difficulty is high

wordle.int.summary %>%
  ungroup() %>%
  mutate(
    difficulty = factor(
      ifelse(attempts_mean <= 4.1, "low", "high"), 
      levels=c("low", "high")
    )
  ) -> wordle.int.summary

Next, we randomly (but reproducibly) partition wordle.int.summary into training and testing sets:

set.seed(1) # for reproducibility
sample <- sample(nrow(wordle.int.summary),floor(nrow(wordle.int.summary)*0.7))
training <- wordle.int.summary[ sample,]
testing  <- wordle.int.summary[-sample,]

Q4.1

Fit a logistic regression model to the training set using the freq, vowels and the repeated as predictors of difficulty

(1 mark)

# Write your answer here   
glm(difficulty ~ freq + vowels + repeated, data=training, family="binomial") -> training.glm
training.glm


Call:  glm(formula = difficulty ~ freq + vowels + repeated, family = "binomial", 
    data = training)

Coefficients:
(Intercept)         freq       vowels     repeated  
    -5.0909      -0.2527       0.4821       1.3258  

Degrees of Freedom: 150 Total (i.e. Null);  147 Residual
Null Deviance:      202.9 
Residual Deviance: 189.8    AIC: 197.8

Q4.2

Predict the difficulty in the testing set and save the results to the testing set as a new variable called prediction

(1 mark)

# Write your answer here   
testing$prediction <- training.glm %>% predict(testing, type="response")
testing$prediction

        1         2         3         4         5         6         7         8         9        10        11        12 
0.6445979 0.5781986 0.3541735 0.5189757 0.7568412 0.5648915 0.4186560 0.4216720 0.4031905 0.6035482 0.4638245 0.6849840 
       13        14        15        16        17        18        19        20        21        22        23        24 
0.8756442 0.2896762 0.7005997 0.6022788 0.6600108 0.3241230 0.6287186 0.4843781 0.6101171 0.7897278 0.5618286 0.4023296 
       25        26        27        28        29        30        31        32        33        34        35        36 
0.6788279 0.8444698 0.7798131 0.5797514 0.5960917 0.5406892 0.6472906 0.2902641 0.6403225 0.6897535 0.6471455 0.4133288 
       37        38        39        40        41        42        43        44        45        46        47        48 
0.5103873 0.8993954 0.5184869 0.7244255 0.6786495 0.7363602 0.7192409 0.4021544 0.3674010 0.6039731 0.5411835 0.8018943 
       49        50        51        52        53        54        55        56        57        58        59        60 
0.4821190 0.7069143 0.4272740 0.6891867 0.8265687 0.7812279 0.4902826 0.5612018 0.7308136 0.6617327 0.5991350 0.8900565 
       61        62        63        64        65 
0.5786064 0.7098222 0.4774360 0.4794829 0.4370228

Q4.3

Apply a decision threshold to testing$prediction to create a new variable called testing$predicted.difficulty whose value of “low” if testing$prediction $< 0.5$ and “high” otherwise. * Print out the first ten values of testing$predicted.difficulty

(1 mark)

# Write your answer here   
testing %>%
  ungroup() %>%
  mutate(
    predicted.difficulty = factor(
      ifelse(prediction < 0.5, "low", "high"), 
      levels=c("low", "high")
    )
  ) -> testing
head(testing$predicted.difficulty, 10)

   1    2    3    4    5    6    7    8    9   10 
high high  low high high high  low  low  low high 
Levels: low high

Q4.4

Using your testing set predictions, demonstrate commands that show the confusion matrix with a decision threshold of 0.5.
- Hint: use table()
- Your output should be:

         actual
predicted low high
     low   13    6
     high  21   25

(1 mark)

# Write your answer here   
confusion_matrix <- table(testing$predicted.difficulty, testing$difficulty)
names(dimnames(confusion_matrix)) <- c("predicted", "actual")
confusion_matrix

         actual
predicted low high
     low   13    6
     high  21   25

Taking “high” to be positive, calculate and print
- a variable called TPR, the true positive rate of your model on the test set
- a variable called FPR, the false positive rate of your model on the test set
  (1 mark)

# Write your answer here   
TP = 25
FP = 21
FN = 6
TN = 13
P = TP + FN
N = TN + FP

TPR = TP/P
FPR = FP/N

TPR

[1] 0.8064516

FPR

[1] 0.6176471

As demonstrated in Workshops, use the simple.roc() function (below) to plot the Receiver Operating Characteristic curve of your model’s predictions on the test set
- annotate it with a red point showing the True Positive Rate and False Positive Rate from the previous question
- your result should look like this

(1 mark)

# Modified from https://blog.revolutionanalytics.com/2016/08/roc-curves-in-two-lines-of-code.html
# labels is a logical vector indicating whether an example belongs to the positive class
# scores is the predicted probability of that example belonging to the positive class 
simple.roc <- function(labels, scores){
  ordered.scores <- order(scores, decreasing=TRUE)
  labels <- labels[ordered.scores]
  tibble(
    TPR=c(0,cumsum( labels))/sum( labels),
    FPR=c(0,cumsum(!labels))/sum(!labels),
    labels=c(NA,labels),
    score=c(Inf, scores[ordered.scores])
  )
}

# Write your answer here   
#difficulty.roc <- simple.roc(testing$predicted.difficulty=="high", )
#diagonal <- data.frame(FPR=c(0,1), TPR=c(0,1))

#ggplot(data=difficulty.roc, aes(x=FPR, TPR)) +
  #coord_equal(xlim=c(0,1), ylim=c(0,1)) +
  #geom_line(data=diagonal, lty=2) +
  #labs(xlab="False Positive Rate (FPR)", ylab="True Positive Rate (TPR)")

# Uncomment to terminate knitting
#knitr::knit_exit()

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❔ Question 5: Ethics in Data Science (3 marks total)

Choose only one of the following two questions to answer.

As a guide, each of your answers should be around 5 lines of text.

Q5.1

Word Tips published some interesting Wordle results comparing the Wordle solving skills of different cities worldwide.

These ranks were obtained based on the number of attempts via Twitter feed results.
This site crowned Canberra as the world’s best Wordle city.

Do you think that this data constitutes a representative sample of Wordle players? Explain your reasoning.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

Considering issues of variation, what would you say to help someone understand the meaning of the averages reported in this article.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

Do you think it would be ethical to make a life-changing decisions about two groups based on comparing their sample means without considering variation? Explain your reasoning.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

Q5.2

Multiple users have tweeted wordle results with a perfect score, i.e., they have correctly guessed the word in one attempt. This seems quite unlikely to happen.

Do you think there are any ethically significant benefits or harms in cheating in a game like this? Justify your answer.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

I believe that is a balanced amount of ethically significant benefits or harms in cheating in games such as Wordle. Generally, playing Wordle does not have an external effect to the people around you irregardless of whether you are playing honestly or dishonestly. Neither completing nor failing to do your Wordle will not bring about any significant benefit to an individual’s life. Though you could bring the argument that ethical practice of following rules and specifications is necessary, and willfully ignoring it is an ethically significant harm. Furthermore, cheating presents inaccuracies and outliers within data that may affect the human understanding and reduces fairness and justice.

Would you cheat at Wordle? Explain your answer in terms of one or more of the three main ethical frameworks presented by Shannon Vallor in Introduction to Data Ethics.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

I would not cheat at Wordle. An ethical framework presented by Shannon Vallor in Introduction to Data Ethics that forms my reasoning is Deontological Ethics, or rule or principle-based systems of ethics. The rules in Wordle does not permit us to cheat, and to play to game truthfully with natural, human mistakes arising. You are not allowed to know of the words or the potential letters before you start playing your daily Wordle. This applies to all games, generally, in which you must not give yourself an unfair advantage over other players to complete the game as this is a breach of morals, truthfulness and the general rules of the game.

Describe another realistic scenario where people may be tempted to cheat or misrepresent the truth, and in doing so, provide misleading data.

Write your answer here as an Rmarkdown blockquote (lines beginning with >):

(1 mark)

There are several different scenarios in which people may be tempted to cheat or misrepresent the truth. One major example of a misrepresentation of truth in recent times are people claiming that various elections have been ‘rigged’ are claim that votes have been miscounted in favour or a different candidate which may not cater to their political compass. Some news outlets have also enabled this, generally in favour of one political party over the other, and misrepresent data in their broadcasts to imply that one candidate is more popular than the other. This misleads the population of just exactly that, and may influence their votes in not only a current election, but also future elections.

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❔ Question 6: Using `R` for data science (2 marks total)

Q6.1

In completing this Assessment, use the principles and approaches espoused in R for Data Science by Garrett Grolemund and Hadley Wickham to ensure that your work can be run, reproduced and understood by the Teaching Team.
(1 mark)

Q6.2

Use your student ID as the name of your working directory
Within your working directory
- Edit your solutions into this file (Assessment02.Rmd)
  - We will run this file to mark your solutions
- Knit your solutions to create the file Assessment02.html in your working directory
Zip your working directory to create a zip file named according to your student ID
Upload your zip file submission via Blackboard/Assessment02
- Please make multiple submissions as evidence of your progress that we can take into account if there are any unexpected issues with your final submission
To illustrate what we expect, if your student ID was n12345678
- You should submit a file called n12345678.zip
- This file should unzip to create a folder called n12345678 which contains the files Assessment02.Rmd and Assessment02.html

(1 mark)

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Assessment02: Wordle

DSB💯 Fundamentals of Data Science

Student name: Ryan Indrananda Student number: 10852565

Semester 2, 2022

🟩 The story of Wordle

Instructions

📢 About this Assessment item

📩 How to submit this Assessment item

😎 How to succeed with this Assessment item

🛠️ Work in a new directory, using relative file names

🏡 Work from home!

🧹 Start with a clean environment

📚 Load only these libraries

🟨 Your datasets

❔ Question 1: Summarizing wordle data (8 marks total)

Q1.1

Q1.2

Q1.3

Q1.4

Q1.5

Q1.6

Q1.7

❔ Question 2: Exploring Australian wordles (6 marks total)

Q2.1

Q2.2

Q2.3

❔ Question 3: Linear regression (5 marks total)

Q3.1

Q3.2

Q3.3

Q3.4

Q3.5

❔ Question 4: Logistic regression (6 marks total)

Q4.1

Q4.2

Q4.3

Q4.4

❔ Question 5: Ethics in Data Science (3 marks total)

Q5.1

Q5.2

❔ Question 6: Using R for data science (2 marks total)

Q6.1

Q6.2

Student name: Ryan Indrananda
Student number: 10852565

❔ Question 6: Using `R` for data science (2 marks total)